My Math Forum The empty set

 Applied Math Applied Math Forum

 September 17th, 2013, 10:46 PM #1 Member   Joined: Sep 2013 Posts: 33 Thanks: 0 The empty set If P(A) is the power set of A=[a,b] and we no that B is a non-empty subset of A does B might be the empty set ? On one hand I just said the B is non-empty, on the other P(A) includes the empty set, so being the empty set B still includes an actual element of P(A). I am confused. Thank you.
 September 18th, 2013, 01:07 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: The empty set First, the empty set is a subset of any set. But, this doesn't mean that it contains any element of the set. Here's an example that may help: Suppose you have a choice of a Coke or a Sprite. You actually have four options: You could drink the Coke. You could drink the Sprite. You could be greedy and drink both! Or, you could drink neither. This last option (not to choose anything that is available) is logically the same as the empty set being a subset of any set.
September 19th, 2013, 06:32 AM   #3
Math Team

Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: The empty set

Quote:
 Originally Posted by barokas If P(A) is the power set of A=[a,b] and we no that B is a non-empty subset of A does B might be the empty set ? On one hand I just said the B is non-empty, on the other P(A) includes the empty set, so being the empty set B still includes an actual element of P(A). I am confused. Thank you.
Your post is confused! "If P(A) is the power set of A=[a,b] and we know that B is a non-empty subset of A does B might be the empty set ?" No, if we "know B is non-empty subset of A" then B is not the empty set!

Had you just said "B is a subset of A" or "B is a member of P(A)" then B might well be empty. Yes, P(A) includes the empty set but that has nothing to do with "includes an actual element of A". (Another confusion- "actual elements of P(A)", which is what you wrote, are subsets of A. "Subsets of A" do NOT generally contain other subsets of A.)

You may be confused about the definition of "subset of A". Saying that "B is a subset of A" does NOT necessarily mean B "includes an actual element of A". B is a subset of A if B does NOT include any element that is NOT in A. Since the empty set does NOT include any elements, it cannot include any that are "not in A" and so is a subset of any set, A.

P(A) is the set of all subsets of A, NOT just "all non-empty subsets of A" and always includes the empty set.

 September 22nd, 2013, 08:32 PM #4 Member   Joined: Sep 2013 Posts: 33 Thanks: 0 Re: The empty set Thanks, my question was indeed confusing as i wanted to ask if b may contain the emty set i learned that the answer isnpositive. Thanks again
September 25th, 2013, 03:47 PM   #5
Math Team

Joined: Apr 2012

Posts: 1,579
Thanks: 22

Re: The empty set

Quote:
 Originally Posted by barokas Thanks, my question was indeed confusing as i wanted to ask if b may contain the emty set i learned that the answer isnpositive. Thanks again
If B is the power set of A, then B not just can but invariably does contain the empty set. You never defined B.

 Tags empty, set

Proving that an emty set is a subset of any set

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Vasily Applied Math 2 August 19th, 2012 12:31 PM jstarks4444 Applied Math 1 October 12th, 2011 05:40 PM outsos Applied Math 36 April 30th, 2010 10:46 AM z0r Applied Math 2 December 6th, 2008 08:58 PM jstarks4444 Number Theory 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top